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As i understand it, this question should be answered by N(A)/N.

I take N to be 10!, im not sure if this is correct.

How do you calculate N(A)?

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If you accept an answer, click the check mark at the upper-left of the answer. It is proper behavior and gives feedback to the people providing assistance. – jrand Dec 9 '12 at 3:23

We can use your $10!$. There certainly are $10!$ equally likely ways to arrange the books.

Now we count the number of orderings in which our $3$ books are together. The leftmost of our $3$ books can be in any of $8$ places, $1$ to $8$. (Write the numbers $!$ to $10$ in a row, and look.) Once this place is chosen, implicitly the other locations are chosen. Then the $3$ books can be put in the places in $3!$ orders. And for each such arrangement, the other books can be chosen in $7!$ ways, for a total of $(8)(3!)(7!)$. Now divide by $10!$. We get probability $\dfrac{(8)(3!)(7!)}{10!}$. There is a lot of cancellation.

The following is perhaps easier. The set of positions our $3$ books will occupy can be chosen in $\dbinom{10}{3}$ equally likely ways.
Note that we are not placing the books in these places yet, we are just choosing a set of locations.

Of these ways, $8$ have our books in consecutive order. So the probability is $\dfrac{8}{\binom{10}{3}}$.

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Thank you so much, the answer helps alot! – John Dec 9 '12 at 2:16

here is another angle to the solution which is correctly given by André Nicolas: take those 3 books as one-- will occupy 1 slot. then you have 8 different items for 8places: 8!. and you can arrange those 3 in 3! ways. multiply.

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